2

I would like to know if there is a (preferably closed-form) solution for

$B \ln y +A y \ln y + A y- A =0$ for $y$

Where $A, B \in \mathbb{R}^{+}$. I have reasons to think there isn't a closed form solution or an approximation that is valid when $r$ is in the order of $B$ and $B \rightarrow 0$.

Is there any theorem saying there are no closed-form solutions to this equation? If there is, do you think an approximate solution might be possible? If there isn't, any hint how to solve it?

Note: My question is motivated by the need for a solution to this problem:

$1-\frac{A}{x} \gamma (2,\frac{x}{B}) = 0$ for $x$

The solution needn't be exact, but it'd be great if it was in closed form (I'm trying to avoid numerical approximations). Especially, I'm interested in the case $r$ around $B$, but also would like convergence for $B \rightarrow 0$.

My approach was to perform an asymptotic expansion on the incomplete lower gamma function following http://dlmf.nist.gov/8.11, truncating the series after 2 terms. After simplifying, Mathematica gives me:

$x + x \frac{A}{B} e^{-\frac{x}{B}}+A e^{-\frac{x}{B}} -A =0$

So, after the substitution $x=B \ln y$, I get: $B \ln y +A y \ln y + A y-A =0$ for $y$

Maybe there is a better approach to this problem?

Thanks!

misi
  • 1,025
  • The original poster has declared (http://math.stackexchange.com/questions/353156/solve-b-ln-y-a-y-ln-y-a-y-a-0-for-y) that this question isn't the intended one. – Greg Martin Apr 06 '13 at 18:36

2 Answers2

3

Since $A>0$ (the only solution for $A=0$ would be $B=0$ and $y$ arbitrary anyway) we can divide by $A$ to obtain $$(y+\frac{B}{A})\ln y = 1-y$$ Note that $y=1$ is certainly a solution to your equation. Furthermore, note that for $y<1$ the left hand side is negative, whereas the right hand side is positive and conversely, for $y>1$ the left hand side is positive, whereas the right hand side is negative. Hence $y=1$ is the only solution to your equation.

Abel
  • 7,452
  • Thanks for your answer! However, the equation was supposed to be: Solve $-B \ln y +A y \ln y - A y-A =0$, as can be seen from the substitution made in the Note. I am sorry, it was late and I didn't see the typo. Should I open a new question with the corrected expression and grant you this one as solved? – misi Apr 06 '13 at 17:44
  • (Note that the sign changes of 2 of the terms leave me with: $y-1=(\frac{B}{A}+y)\ln y$ and I can no longer use your argument) – misi Apr 06 '13 at 17:46
-1

I doubt that the equation has a nontrivial solution. Implicitly differentiating once,

$$ B/y + 2A + A\ln y = 0. $$

Rearranging,

$$ \mathrm{e}^2y = \mathrm{e}^{-B/Ay}. $$

Implicitly differentiating once more,

$$ \mathrm{e}^2 = -\frac{B}{A}\mathrm{e}^{-B/Ay} $$

which seems to suggest that $y$ is constant.

Herng Yi
  • 3,236
  • Implicit differentiation requires one to deal with the derivatives like $\dfrac{\mathrm dB}{\mathrm dy}, \dfrac{\mathrm dA}{\mathrm dy}$ or their inverses; if we assume these are zero it's no surprise that $y$ in terms of $A, B$ can only be a constant. – Lord_Farin Apr 06 '13 at 07:48
  • @Lord_Farin $A$ and $B$ are constants, as the OP writes in his post, so these derivatives will be zero. – Elmar Zander Apr 06 '13 at 07:51
  • @Elmar But if it were that $y = AB$, say, then $\dfrac{\mathrm dy}{\mathrm dB} = A$ so $\dfrac{\mathrm dB}{\mathrm dy} = 1/A$ regardless of whether $A,B$ are constant. What the answer obtains is that $y$ is constant WRT some non-present variable $t$, say. This is useless information. – Lord_Farin Apr 06 '13 at 07:54
  • @Lord_Farin how am I depending on an external variable? I only differentiate with respect to $y$, so that all of the statements in my answer should be true for all positive $y$ (due to the use of $\ln$). That would make my conclusion inevitable. – Herng Yi Apr 06 '13 at 08:07
  • Since $y$ will be an expression in terms of $A, B$, it is unfounded to assume the differential quotients I put in my earlier posts are zero. Try your approach on $y - B = 0$. Admittedly this probably has nothing to do with an external variable; that was probably a quirk of my brain - sorry for that. – Lord_Farin Apr 06 '13 at 08:10
  • If I assume that $B$ is constant in $y - B = 0$, then to implictly differentiate I first assume that $y$ is not constant (or differentiation would not be defined). I then differentiate to get $1 = 0$, which is absurd so apparently $y$ has to be a constant. – Herng Yi Apr 06 '13 at 08:26